aqa c2 revision notes
DESCRIPTION
AQA C2 Maths Revision Notes (As Level)TRANSCRIPT
CORE 2 Summary Notes
1 Indices! Rules of Indices
xa ¥ x
b = xa + b
xa $ x
b = xa – b
(xa)b = xab
Solve 32x + 1 – 10 ¥ 3x + 3 = 0
3 ¥ (3x)2 – 10 ¥ 3x + 3 = 0
Let y = 3x
3y2 – 10y + 3 = 0
(3y – 1)(y – 3) = 0
y = 1 gives 3x = 13 3
x = – 1
y = 3 gives 3x = 3 x = 1
Solve 9x = 127
32x = 3 – 3
2x = – 3
x = – 32
! x0 = 1
!x
– a = 1a
x
! xn = n1
x
! x n n m n mm
x)x( ==
2 Graphs and TransformationsASYMPTOTES
Straight lines that are approached by a graph which never actually meets them.
y
x1 2 3 4 5– 1– 2– 3– 4– 5
1
2
3
4
5
– 1
– 2
– 3
– 4
– 5
Horizontal Asymptotey = 1
Vertical Asymptotex = 2
y = 1x – 2
+ 1
TRANSLATION - to find the equation of a graph after a translation of ˙ you replace x
by (x-a) and y by (y - b)˚
˘ÍÎ
Èab
1
y
x1 2 3 4 5– 1– 2– 3– 4– 5
1
2
3
4
5
– 1
– 2
– 3
– 4
– 5
e.g. The graph of y = x2 -1 is translated through Í ˙ . Write down the equation of˚
˘
Î
È 3- 2
the graph formed.(y + 2) = (x-3)2 – 1
y = x2 – 6x + 6
y - b = f(x-a)
or
y = f(x-a) +by = x2 – 6x + 6
y = x2 -1
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y = -f(x)REFLECTINGReflection in the x-axis, replace y with –yReflection in the y-axis, replace x with – x
STRETCHING
Stretch of factor k in the x direction replace x by 1 xk
Stretch of factor k in the y direction replace y by 1 yk
y = f(-x)
y = f( 1 xk
)
e.g. Describe a stretch that will transform y = x2 + x -1 to thegraph of y = 4x2 + 2x -1
4x2 + 2x -1 = (2x)2 + (2x) -1So x has been replaced by 2x.
Stretch of scale factor ½ in the x direction
y = kf(x)
3 Sequences and Series 1! A sequence can be defined by the nth term such as un = n2 +1
u1 = 12 +1 u2 = 22 +1 u3 = 32 +1 u4 = 42 +1= 2 = 5 =10 =17
! An INDUCTIVE definition defines a sequence by giving the first term and a rule to findthe next terms.
Un+1 = 2un + 1 u1 =3 u2 =7 u3 =15
! Some sequences get closer and closer to a value called the LIMIT – these are knownas CONVERGING sequences
e.g The sequence defined by Un+1 = 0.2un + 2 u1 =3 converges to a limit l
Find the value of l.
l must satisfy the equation I = 0.2l + 2
0.8l = 2
l = 2 ÷ 0.8
= 2.5
ARITHMETIC SEQUENCEEach term is found by adding a fixed number (COMMON DIFFERENCE) to theprevious one.
u1 = a u2 = u1+ d u3 = u2 + d
a is the first term ,d is the common difference the sequence isa, a + d , a + 2d, a + 3d,……….
Un = a + (n - 1)d
SUM of the first n terms of an AP (Arithmetic progression)
Sn = 21 n ( 2a + (n-1)d ) or
21 n(a + l) where l is the last term
! The sum of the first n positive integers is 12
n(n + 1)
! SIGMA notation Si = 0
5
i3 = 03 + 13 + 23 + 33 + 43 + 53
2
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Show that Sr = 1
20
(3n – 1) = 610
= 2 + 5 + 8 + .········59
a = 2 d = 3
S20 = 202
(2 ¥ 2 + (20 – 1) ¥ 3)
= 10 ¥ 61
= 610
4 Geometric Sequences! A geometric sequence is one where each term is found by MULTIPLYING the
previous term by a fixed number (COMMON RATIO)
! The nth term of a geometric sequence a, ar, ar2, ar3, …is arn-1
a is the first term r is the common ratio
! The sum of a geometric sequence a + ar + ar2 + ar3 ….+ arn-1 is a geometric
seriesr 1
1)= a(rn -Sn -
! If -1 < r < 1 then the sum to infinity is1 r
a-
Find the sum to inifinty of the series 1 + 23
+ 49
+ 827
Geometric Series first term = 1 common ratio = 23
Sum to infinity = 1= 3
31-
2
5 Binomial Expansion! The number of ways of arranging n objects of which r are one type and (n-r) are
another is given byr!(n r)!
n!rn
=-˜̃
¯
ˆÁÁË
Ê where n! = n(n-1)(n-2)….x 3 x 2 x 1
! (1 + x)n = 1 + Ê
ËÁ
n ˆ1 ¯
˜ x +Ë
ÊÁ
n ˆ2 ¯
˜ x2 + Ê
ËÁ
n ˆ3 ¯
˜ x3 + .········.xn
(a + b)n = an +Ë
ÊÁ
n ˆ1 ¯
˜ an – 1b1 + Ê
ËÁ
n ˆ2 ¯
˜ an – 2b2 + .··········. Ê
ËÁ
nn – 1 ¯
ˆ 1˜ a bn – 1 + bn
Find the coefficient of x3 in the expansion of (2+3x)9
n = 9 , r = 3 ÁÁÊ 9 (3x)3(2)6
3 ˜̃¯
ˆ
Ë
Note 3 + 6 = 9 (n)
= 84 × 27 x3 × 64 = 145152 x3
3
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6 Trigonometry
! Exact Values – LEARN
45 30 60
sin 45 = 12
cos 45 = 12
tan 45 = 11
1¥�
45
sin 30 = 12
cos 30 = 32
tan 30 = 13
1 1
2 2sin 60 = 3
2
cos 60 = 12
tan 60 = 3
30
60¥�
COSINE RULE
ab
c
A
B
C
b
a2 = b
2 + c2 – 2bc cos A
2 = a2 + c
2 – 2ac cos B
c2 = a
2 + b2 – 2ab cos A
SINE RULEa
sinA
= b
sinB
= c
sinC
AREA of a TRIANGLE1
ab sinC = 12 2
bc sinA = 12
ac sinB
e.g. Find the area of triangle PQR
Finding angle P
82
= 92
+ 122
– 2 ¥ 9 ¥ 12 ¥ cos P
cos P =9
2
+ 122
– 82
2 ¥ 9 ¥ 12
= 0·745
P = 41·8·····.
Area = 12
¥ 9 ¥ 12 ¥ sin 41·8·····.
= 35·99991 36 cm2 =
= 36 cm2
9cm
12 cm
8 cm
R
P Q
Try to avoid rounding until you reach your finalanswer
180 = p radians 90 = p2
radians 60 = p3
radians
45 = p4
radians 30 = p6
radians
RADIANS and ARCS������ ��ʌ�UDGLDQV
4
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/(1*7+�2)�$1�$5&� O� �Ușl
AREA of SECTOR = ½ r2ș
r
r
7 Trigonometry
\� �VLQ�ș
VLQ���ș�� ���VLQ�ș� VLQ������±�ș�� �VLQ�ș
y
x90 180 270 360– 90– 180– 270
1
– 1
VLQ��ʌ�±�ș�� �VLQ�ș� �LQ�UDGLDQV�
\� �FRV�ș
FRV���ș�� �FRV�ș� FRV������±�ș�� ���FRV�ș
y
x90 180 270 360– 90– 180– 270
1
– 1
FRV���ʌ�±�ș�� ���FRV�ș��LQ�UDGLDQV�
\� �WDQ�șy
x90 180 270 360– 90– 180– 270
5
10
– 5
– 10
tan x = sinx
cos x
7DQ���ș�� ���WDQ�ș
sin2x + cosx
2 = 1LEARN THIS IDENTITYSOLVING EQUATIONS
e.g. Solve the equation 2sin2 [� ���FRV�[�IRU�����[�����ʌ
2
2 sin2x = 3 cosx
2(1 – cos2x) = 3 cos x
cos2x + 3cosx – 2 = 0
(2cos x – 1)(cos x + 2) = 0
cos x = -2 has no solution
5
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cos x = -½ x = p3
or 2 p -3
p = 5p
3
PROVING IDENTITIESShow that (sin x + cos x )2 = 1 + 2sinx cos x
LHS: (sin x + cos x)2 = (sin2 x + 2 sinx cos x + cos2
x)
= sin2 x + cos2
x + 2 sinx cos x
= 1 + 2 sin x cos x
LHS = RHS
TRANSFORMATIONSStretch in the y direction�VFDOH�IDFWRU�D� \� �D�VLQ�ș
6
y = sin x y = sin (x+30)y
x90 180 270 360– 90– 180
1
2
– 1
– 2
A
Stretch in the ș�GLUHFWLRQ scale factor 1b
\� �FRV� Eș
y = sin x y = 4 sin xy
x90 180 270 360– 9080– 1
12345
– 1– 2– 3– 4
\� �FRV�ș� \� �FRV��șy
x90 180 270 360– 90– 180
1
2
– 1
Translation ÍÈ \� �VLQ��ș���F�˙
˚
- c˘
Î 0
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Translation ÍÈ
˙� \� �FRVș���G˚
0˘
Îd\� �FRV�ș� \� �FRVș���
y
x90 180 270 360– 90– 180
12345
– 1
A
8 Exponentials and Logarithms! A function of the form y = ax is an exponential function
! The graph of y = ax is positive for all valuesof x and passes through the point (0,1)
! A Logarithm is the inverse of anexponential function
y = ax fi x = log
a y
e.g log2 32 = 5 because 25 = 32
LEARN THE FOLLOWING
loga a = 1 log
a 1 = 0
loga a
x = x alog x = x
Laws of Logarithmslog
a m + log
a n = log
a mn
loga m – log
a n = log
a
Ê m
ËÁ
n ¯
ˆ˜
kloga m = log
a m
k
Solve logx 4 + log
x 16 = 3
logx 64 = 3
x3 = 64
x = 4
y = 3 – x y = 4x
y = 2x
y
x1 2 3– 1– 2– 3
2
4
6
8
10
12
14
16
E.g
7
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! An equation of the form ax =b can be solved by taking logs of both sides
9 Differentiation and Integration
! If y = xn then dy = nx
dx
n – 1
! Ú xn dx = x
n + 1
n + 1+ c for all values of n except n = – 1
e.g Ú 1 + x
2
x
= Ú x –
12 + x
32 = 2x
12 + 2
5x
52 + c
TRAPEZIUM RULEThe trapezium rule gives an approximation of the area under a graph.If the gap between the ordinates is h then
Area = ½ h (end ordinates + twice ‘sum of interior ordinates’)
Use the trapezium rule 4 strips to estimate the area under the graph of y = 1 + x2
from x = 0 to x = 2
x y0 1���� ¥������ ¥����� ¥����
y
x1 2 3– 1
1
2
– 1
Area =½ x 0.5 ( 1 + ¥���������¥�������¥���
= 1.966917= 1.97 (3sf)
8
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